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Let us consider two spaces $\mathbb{X},\mathbb{Y}$. For simplicity we put $\mathbb{X} = \mathbb{Y} = \mathbb{R}$. On the product space $\mathbb{S} = \mathbb{X}\times \mathbb{Y}$ we consider a Markov process $S = (X,Y)$ with a transition kernel $T$ such that $$ T(B|s) = \mathsf P\{S_1\in B|S_0 = s\}. $$

For any fixed $y\in\mathbb Y$ we construct a Markov process $X^y$ on $\mathbb{X}$ such that its transition kernel $T^y$ is given by $$ T^y(A|x) = T(A\times \mathbb Y|(x,y)). $$

For a set $A\subset \mathbb X$ we are interested in the following probability $$ u(s) = \mathsf P\{S_n\in A\times \mathbb Y\text{ for all }n\geq 0|S_0 = s\}. $$

What I have is that for all $y\in \mathbb Y$ and $x\in\mathbb X$ holds $$ \mathsf P\{X^y_n\in A\text{ for all }n\geq 0|X_0 = x\} = 0. $$

Can I derive from here that $u(s)\equiv 0$ or if there is a counterexample?

It will be even helpful if $\mathbb{X},\mathbb{Y}$ are finite (I mean for the proof or a counterexample).

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up vote 2 down vote accepted

Let $\mathbb{X}=\mathbb{Y}=\mathbb{R}$. Suppose that $$ P(S_1\in B\mid S_0=(x,y)) = \delta_{(x-2^{-y},y+1)}(B). $$ (Here, $\delta_z$ will always denote the point mass at $z\in\mathbb{R}^n$.) In other words, the process $S$ is a (deterministic) Markov chain that goes from $(x,y)$ to $(x-2^{-y},y+1)$ with probability one. In this case, \begin{align*} P(X^y_1\in A\mid X^y_0=x) &= P(S_1\in A\times\mathbb{R} \mid S_0=(x,y))\\ &= \delta_{(x-2^{-y},y+1)}(A\times\mathbb{R})\\ &= \delta_{x-2^{-y}}(A). \end{align*} In other words, $X^y$ is a (deterministic) Markov chain that goes from $x$ to $x-2^{-y}$ with probability one.

Now let $A=(0,\infty)$. For all $x,y\in\mathbb{R}$, as soon as $n\ge x2^y$, we will have $X^y_n = x - n2^{-y} \le 0$, and hence, $X^y_n\notin A$. It follows that $$ P(X^y_n\in A\text{ for all $n\ge0$}\mid X_0=x) = 0. $$ On the other hand, if $S_0=(x,y)$, then $$ S_n = \bigg(x-2^{-y}\sum_{j=0}^{n-1}2^{-j},y+n\bigg). $$ Letting $n\to\infty$, we see that the first component of $S_n$ will be positive for all $n$ if and only if $x-2^{-y+1}\ge0$. Thus, any $s=(x,y)$ which satisfies $x\ge2^{-y+1}$ (for example, $s=(1,1)$), will also satisfy $$ P(S_n\in A\times\mathbb{R}\text{ for all $n\ge0$}\mid S_0=s) = 1. $$ Informally speaking, the process $(X_n^y,y)$ (where $y$ is fixed) is constructed by taking $S_n$ and just forcing its second coordinate back to $y$ after each step. The process $S_n$, on the other hand, is allowed to wander freely in the $y$-direction according to the original dynamics. Through this wandering, we are able to produce different asymptotics, even in the $x$-direction, for the two processes $(X_n^y,y)$ and $S_n$.

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